Question

(a) Use data from Appendix D to calculate $\mathbf{∆}{\mathit{H}}^{\mathbf{°}}$and$\mathbf{∆}{\mathit{S}}^{\mathbf{°}}$at${\mathbf{25}}^{\mathbf{°}}\mathit{C}$for the reaction

$\mathbf{2}{\mathbf{\text{CuCl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\rightleftarrows }\mathbf{2}\mathbf{\text{CuCl}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{\text{Cl}}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

(b)Calculate at$\mathbf{∆}{\mathit{G}}^{\mathbf{°}}$$\mathbf{590}\mathit{K}$, assuming$\mathbf{∆}{\mathit{H}}^{\mathbf{°}}$and$\mathbf{∆}{\mathit{S}}^{\mathbf{°}}$are independent of temperature.

(c) Careful high-temperature measurements show that when this reaction is performed at$\mathbf{590}\mathit{K}$$\mathbf{,}\mathbf{∆}{{\mathbf{H}}_{\mathbf{590}}}^{\mathbf{°}}$is$\mathbf{158}\mathbf{.}\mathbf{36}\mathbf{}\mathbf{\text{kJ}}$and$\mathbf{∆}{{\mathit{S}}^{\mathbf{°}}}_{\mathbf{590}}$is$\mathbf{177}\mathbf{.}\mathbf{74}\mathbf{}\mathbf{\text{J}}\mathbf{}{\mathbf{\text{K}}}^{\mathbf{-}\mathbf{1}}$. Use these facts to compute an improved value of$\mathbf{∆}{{\mathit{G}}^{\mathbf{°}}}_{\mathbf{590}}$for this reaction. Determine the percentage error in$\mathbf{∆}{{\mathit{G}}^{\mathbf{°}}}_{\mathbf{590}}$that comes from using the$\mathbf{298}\mathit{K}$values in place of$\mathbf{590}\mathbf{-}\mathit{K}$values in this case.

Short answer

For the reaction the enthalpy$∆H^{°}$ is$165.8\text{kJ}$ and entropy $∆S^{°}$is$177.96\text{J}{\text{K}}^{-1}$ . The$∆G^{°}$ is$270.80\text{kJ}$and error percentage is $2.88\%$.

Step-by-step solution

Given information

The reaction of Copper(II) chloride at${25}^{°}C$is given below:

$2{\text{CuCl}}_2\left(s\right)\to 2\text{CuCl}\left(s\right)+{\text{Cl}}_2\left(g\right)$

Concept of Gibb’s free energy

Gibbs free energy, also known as Gibbs function, Gibbs energy, or free enthalpy, is a term used to estimate the greatest amount of work done in a thermodynamic system when temperature and pressure remain constant. Therefore, the difference between the system's temperature and entropy change and the change in enthalpy equals the change in Gibbs free energy. The second law of thermodynamics states that for a spontaneous process, entropy in the universe always increases. G determines the direction and magnitude of chemical change.

Calculate the enthalpy and entropy

The standard enthalpies:

$$\begin{gathered} \Delta H_f^{°}\left({\text{CuCl}}_2\right)=-220.1\text{kJ} \\ \Delta H_f^{°}\left({\text{Cl}}_2\right)=0\text{kJ} \\ \Delta H_f^{°}\left(\text{CuCl}\right)=-137.2\text{kJ} \end{gathered}$$

Calculate the change in enthalpy:

$\Delta H°=\sum \Delta H_f^{°}\left(\text{products}\right)-\sum \Delta H_f^{°}\left(\text{reactants}\right)$

For the reaction formula would be:

$\Delta H°=2\Delta H_f^{°}\left(\text{CuCl}\right)+\Delta H_f^{°}\left({\text{Cl}}_2\right)-2\Delta H_f^{°}\left({\text{CuCl}}_2\right)$

 Step 4: Calculate the change in enthalpy

Calculate the change in enthalpy

$$\begin{gathered} \Delta H^{°}=2\times (-137.2)+0-2\times (-220.1) \\ \Delta H^{°}=165.8\text{kJ} \end{gathered}$$

The standard entropies:

$$\begin{gathered} ∆S_f^{°}\left({\text{CuCl}}_2\right)=86.2\text{J}{\text{K}}^{-1} \\ ∆S_f^{°}\left({\text{Cl}}_2\right)=222.96\text{J}{\text{K}}^{-1} \\ ∆S_f^{°}\left(\text{CuCl}\right)=108.7\text{J}{\text{K}}^{-1} \end{gathered}$$

Calculate the ∆G°

Calculate the$∆G°$at$590K$assuming$∆H°$and$∆S°$are independent of temperature.

The formula for the change in Gibb's energy:

$\Delta G^{°}=H°+T\Delta S°$

Then, substitute the values from above:

$$\begin{gathered} \Delta G°=165.8\times {10}^3+590\times 177.96 \\ \Delta G°=270.80\text{kJ} \end{gathered}$$

Calculate the error percentage

Calculate the percentage error of $∆G_{590K}^{°}$using the$298k$values of the change in enthalpy and entropy if the experimental data is:

$$\begin{gathered} \Delta H°=158.36\text{kJ} \\ \Delta S°=177.74\text{J}{\text{K}}^{-1} \end{gathered}$$

Then calculate the actual change in Gibb's free energy at $590k$using the same formula as:

$$\begin{gathered} \Delta G°=\Delta H°+T\Delta S° \\ \Delta G°=158.36\times {10}^3+590\times 177.74 \\ \Delta G°=263.23\text{kJ} \end{gathered}$$

$$\begin{gathered} \Delta G°=\Delta H°+T\Delta S° \\ \Delta G°=158.36\times {10}^3+590\times 177.74 \\ \Delta G°=263.23\text{kJ} \end{gathered}$$

Finally, calculate the error percentage:

$$\begin{gathered} \text{error}=\frac{\Delta G_C^{°}-\Delta G_C^{°}}{\Delta G_C^{°}}\times 100\% \\ \text{error}=\frac{270.80-263.23}{263.23}\times 100\% \\ \text{error}=2.88\% \end{gathered}$$